## Calculus: Early Transcendentals 8th Edition

$(\dfrac{\partial z }{\partial x})(\dfrac{\partial x }{\partial y})(\dfrac{\partial y}{\partial z})=-1$
Recall Equation 7, which is: $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$ $(\dfrac{\partial z }{\partial x})(\dfrac{\partial x }{\partial y})(\dfrac{\partial y}{\partial z})=- \dfrac{F_x}{F_y}\dfrac{F_y}{F_x}\dfrac{F_z}{F_y}$ $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$ This gives: $(\dfrac{\partial z }{\partial x})(\dfrac{\partial x }{\partial y})(\dfrac{\partial y}{\partial z})=-1$